Absorption 6c396s

Absorption and Stripping  Absorption (or scrubbing) is the removal of a component (the solute or absorbate) from a gas stream via uptake by a nonvolatile liquid (the solvent or absorbent).  Desorption (or stripping) is the removal of a component from a liquid stream via vaporization and uptake by an insoluble gas stream.  Thus, absorption and stripping are opposite unit operations, and are often used together as a cycle.  Both absorption and stripping can be operated as equilibrium stage processes using trayed columns or, more commonly, using packed columns. Lecture 21

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In absorption (also called gas scrubbing and gas washing), a gas mixture is ed with a liquid (absorbent or solvent) to selectively dissolve one or more components by mass transfer from the gas to the liquid. The components transferred to the liquid are referred to as solutes or absorbate.

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The opposite of absorption is stripping (also called desorption) where in a liquid mixture is ed with a gas to selectively remove components by mass transfer from the liquid to the gas phase. Absorbers are frequently couples with strippers to permit regeneration and recyling of absorbent. When water is used as the absorbent, it is common to separate the absorbent from the solute by --------------

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Absorption and Stripping

When molasses is fermented to produce a liquor containing ethyl alcohol a carbon dioxide rich vapor containing a small amount of ethyl alcohol is evolved. The alcohol can be recovered by adsorption with water in a sieve tray tower. For the following conditions determine the number of equilibrium stages required for counter current flow of liquid and gas, assuming isothermal, isobaric conditions in tower and neglect mass tranfer of all components except ethyl alcohol.

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Absorber/Stripper Cycle

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Absorption Systems – Physical 

Examples: CO2 and water CO and water H2S and water NH3 and water NO2 and water

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Physical absorption relies on the solubility of a particular gas in a liquid.

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This solubility is often quite low; consequently, a relatively large amount of liquid solvent is needed to obtain the required separation.

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This liquid solvent containing the solute is typically regenerated by heating or stripping to drive the solute back out.

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Because of the low solubility and large solvent amounts required in physical absorption, chemical absorption is also used…

Acetylene and acetic acid NH3 and acetone Ethane and carbon disulfide N2 and methyl acetate NO and ethanol

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Absorption Systems – Chemical 

Chemical absorption relies on reaction of a particular gas with a reagent in a liquid.

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Examples:

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This absorption can often be quite high; consequently, a smaller amount of liquid solvent/reagent is needed to obtain the required separation.

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However, the reagent may be relatively expensive, and it is often desirable to regenerate when possible.

CO2 / H2S and aqueous ethanolamines CO2 / H2S and aqueous hydroxides CO and aqueous Cu ammonium salt SO2 and aqueous dimethyl aniline HCN and aqueous NaOH HCl / HF and aqueos NaOH

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Absorption and Stripping – the Problem 

The principal difference in handling adsorption and stripping, compared to distillation, is how we represent the equilibria (equilibrium curve) and mass balances (operating lines).

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In distillation, the liquid and vapor streams were assumed to remain constant under CMO.

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In absorption, we have removal of the solute from the gas stream and uptake by the solvent liquid stream; thus, the total liquid and gas stream amounts or flow rates can change.

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If we use mole fractions of the solute and assume that the gas and liquid stream amounts or flow rates remain constant, significant error can result if the solute concentration in the inlet gas stream is greater than about 1%.

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If we can set up our equilibrium curve and operating line to for this change in the overall gas and liquid flow rates, we can use the McCabe-Thiele method to solve absorption and stripping problems. Lecture 21

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Absorption and Stripping Assumptions We assume that:

 The carrier gas is insoluble (or it has a very low solubility), e.g, N2 or Ar in water.  The solvent is nonvolatile (or it has a low vapor pressure), e.g., water in air at low temperatures.  The system is isothermal. e.g., the effects of heat of solution or reaction are low or there is a cooling or heating system in the column.  The system is isobaric.  The concentration of the solute is low, say <10% – this is the limit for the use of Henry’s Law, which we will discuss later. Lecture 21

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General Design consideration

Entering gas (liquid) flow rate, composition, temperature and pressure. Desired degree of recovery of one or more solutes. Choice of absorbent (stripping agent) Operating pressure and temperature, and allowable gas pressure drop. Minimum absorbent(stripping) flow rate and actual absorbent(stripping) flow rate as a multiple of the minimum rate needed to make the seperation. Number of equilibrium stages and stage efficiency Heat effects and need for cooling(heating). Type of absorber(stripper) equipment. Height of absorber(stripper). Diameter of absorber(stripper). Lecture 21

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Characterastics of a ideal absorbent

Ideal absorbent should have high solubility of the solute Have a low volatility to reduce the loss of absorbent and facilitate seperation of absorbent from solute. Stable to maxmize absorbent life and reduce absorbent make up requirement Have a low viscosity to provide low pressure drop and high mass and heat transfer rates.

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Absorption and Stripping – Circumventing the Problem 

While the total gas and liquid streams can change in absorption, the flow rate of the carrier gas, which we assume to be insoluble in the solvent, does not change.

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Similarly, the flow rate of the solvent, which we assume to be nonvolatile, does not change.

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Consequently, we can define our equilibrium curve and operating line in of mole ratios with respect to the carrier gas and solvent, instead of mole fractions as we did in distillation.

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Doing so circumvents the problem of the changing total gas and liquid stream amounts or flow rates in absorption and stripping.

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Absorption – Variable Specification G, Y1

Total Xo, L’

Total V’, Y1

1

1

→ N

N Total V’, YN+1

L’, X0

GN+1, YN+1

Total L’, XN

Total Vapor (V) and Total Liquid (L) Flows Rates and Mole Fractions

L’, XN

Carrier Gas (G) and Solvent (L) Flow Rates and Mole Ratios

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Liquid Solvent to Carrier Gas Ratio Defining the L and G flow rates as L’ = molar flow rate of nonvolatile solvent (solute-free absorbent) G = molar flow rate of insoluble carrier gas (solute free gas) the assumptions of a nonvolatile solvent and insoluble carrier gas yields L N  L j  L 0  L  constant

Eq. (15-5)

G N 1  G j  G 1  G  constant

Eq. (15-6)

We can define the ratio of liquid to gas as

L' moles nonvolatile solvent/hr   constant G moles insoluble carrier gas/hr Lecture 21

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For application to absorber

L’- molar flow rate of solute free absorbent V’- molar flow rate of solute free carries gas(carrier gas). X mole ratio of solute to solute free absorbent in the liquid Y mole ratio of solute to solute free gas in the vapor

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Mole Ratios We can then define the gas and liquid molar ratios as Y

moles solute in gas moles solute free carier gas

X

moles solute in liquid moles solute free absorbent

Eq. (15-7a)

The molar ratios are related to the mole fractions for solute i by Yi 

yi 1 - yi

Xi 

xi 1- xi

Eq. (15-7b)

K=yn/xn=(Yn/(1+Yn))/((Xn/1+Xn))

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Absorber Operating Line G, Y1

L0, X0 1

N LN, XN

GN+1, YN+1

Yj1 

L L   X j   Y1  X 0  G G   Lecture 21

Eq. (15-9) 23

Henry’s Law – Mole Fraction Relationship  Absorption data is typically available in the form of solute mole fractions, yi vs. xi, or in of the Henry’s constant, H.  Henry’s Law, in of the mole fractions of solute i and the total pressure, is:

Hi yi  xi PTot

Hi yi  PTot x i

 Henry’s Law is valid at low concentrations of solute i, approximately less than 10%. Lecture 21

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Henry’s Law Constants

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Henry’s Law – Mole Ratio Relationship We can rewrite Henry’s Law for solute i, yi 

Hi xi PTot

Eq. (1-3)

in mole ratios, using the mole fraction relationships, Yi 

yi 1 - yi

Xi 

xi 1- xi

Eq. (15-7b)

to yield Henry’s Law in of the molar ratios, Yi and Xi, or Yi H Xi  i 1  Yi PTot 1  X i Lecture 21

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Equilibrium Curve – Mole Ratios Solving for Yi yields Yi 

Xi  Hi   PTot

1

   X i 

We can use this gas molar ratio relationship with the liquid molar ratio Xi 

xi 1- xi

to generate molar ratio equilibrium curves of Yi vs Xi for solute i from xi mole fraction values.

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McCabe-Thiele Plot – Absorber

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Notes on Absorbers 

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Note that the operating line for an absorber is above the equilibrium curve. For a given solute concentration in the liquid, the solute concentration in the gas is always greater than the equilibrium value, which provides the driving force for the separation. The solute is transferred from the gas to the liquid in absorption. In distillation, we plotted the more volatile component, which was transferred from the liquid to the gas. In distillation, if we had plotted the less volatile component, which was transferred from the gas to the liquid, the OL would also lie above the equilibrium curve. Also note that the OL is linear. This results because of the form of the operating line where L/G is a constant. L and G are based upon the nonvolatile solvent and insoluble carrier gas, respectively, which do not change. If we had used mole fractions and total gas and liquid rates, the OL would be curved because the total gas and liquid rates would change since we are removing the solute from the gas and absorbing it into the liquid. One could use mole fractions and the total gas and liquid streams in our calculations only if the solute is in low concentrations, say < 1%, in most systems. Don’t confuse this requirement with that for the use of Henry’s Law, which requires low solute concentrations, < 10%, to be valid. Lecture 21

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Further Notes on Absorbers  Note the location of the top and bottom of the column on the McCabe Thiele diagram.

 We will typically step down from the top of the column, which is equivalent to stepping up on the McCabe-Thiele plot for absorption.  Since we are starting on the OL, we need to express the equilibrium curve in of XEq = XEq(YEq). Although this appears to be opposite of what we did in distillation, where we used yEq = yEq(xEq) when stepping up the plot, that we are still stepping down from the top of the column on the absorption plot.  We use YEq vs. XEq to plot the equilibrium curve, but we also need XEq vs YEq to actually solve the problem analytically. Lecture 21

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Minimum Absorbent Rate – Lmin

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McCabe-Thiele Plot – Stripper

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Final Notes on Strippers    

   

We use the same assumptions and mole ratio methods that we used for adsorbers for the design of strippers. The OL will be the same as that used for absorbers. The difference, compared to an absorber, is that the equilibrium curve will be below the operating line. This is analogous to the stripping section of a distillation column. Just as we stepped up from the bottom of a distillation column’s stripping section, we step up from the bottom of the stripper. Thus, one uses the same McCabe-Thiele algorithm method that we used for stepping up in distillation. In the algorithm, the equilibrium relationship is expressed as YEq = YEq(XEq), and the operating line is expressed in the form YOL = XOL(YOL). Otherwise, the design approach for strippers is the same as that for absorbers. Lecture 21

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When molasses is fermented to produce a liquor containing ethyl alcohol, a CO2 rich vapor containing a small amount of ethyl alcohol is evolved. The alcohol can be recovered by absorption with water in a sieve tray tower. For the following conditions, determine the number of equilibrium stages required for counter current flow of liquid and gas, assuming isothermal, isobaric conditions in the tower and neglecting mass transfer of all components except ethyl alcohol. Entering gas: 98% CO2, 2% ethyl alcohol, 30 deg , 111Kpa. K=0.57 Entering liquid absorbent:100% water, 30 deg C, 110Kpa. Required recovery(absorption) of ethyl alcohol: 97%. Lecture 21

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The exit gas from an alcohol fermenter consists of an air CO2 mixture containing 10% CO2 that is to be absorbed in a 5.0 N solution of triethanolamine containing 0.04 mol of carbon dioxide per mole of amine solution. If the column operates isothermally at 25 deg C, if the exit liquid contains 78.4% of the CO2 in the feed gas to the absorber and if absorption is carried out ina 6 theoretical, plate column calculate: a) Moles of amine solution required per mole of feed gas. b) Exit gas composition. Eqlm data: Y 0.003 0.008 0.015 0.023 0.032 0.043 X 0.01 0.02 0.03 0.04 0.05 0.06 Y 0.055 0.068 0.083 0.099 0.12 X 0.07 0.08 0.09 0.10 0.11 Lecture 21

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Ninety five percent of the acetone vapor in an 85 vol% air stream is to be absorbed by counter current with pure water in a valve tray column with an expected overall tray efficiency of 50%. The column will operate essentially at 20 deg C and 101 Kpa pressure. Equilibrium data for acetonewater at these conditions are: Mole percent acetone in water: 3.30 7.20 11.7 17.1 Acetone partial pressure in air, torr: 30.00 62.80 85.4 103.00 Calculate a) The minimum L/V, the ratio of moles of water per mole of air. b) The number of equilibrium stages required using a value of L/V of 1.25 times the minimum. Lecture 21

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A straw oil used to absorb benzene from coke overn gas is to be stream stripped in a sieve plate column at atmospheric pressure to recover the dissolved benzene. Equilibrium conditions at the operating temperature are approximated are approximated by Henry’s law such that when the oil phase contains 10mol% C6H6, the C6H6 partial pressure above the oil is 5.07 kPa. The oil may be considered non volatile. The oil enters containing 8% benzene, 75% of which is to be recovered. The steam leaving contains 3 mol% C6H6. a) How many theoretical stages are required. B) How many moles of steam are required per 100 mol of oil benzene mixture. C) If 85 % of the benzene is to recovered with the same oil and steam rates, how many theoretical stages are required. Lecture 21

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